Solving a Parallel-Plate Air-Filled Capacitor Problem

[mr_coffee]

Hello everyone. I can't get the 2nd to last question on this problem...
It says:
A parallel-plate air-filled capacitor having area 38 cm2 and plate spacing 2.0 mm is charged to a potential difference of 450 V. Find the following values.
(a) the capacitance
correct check mark pF
(b) the magnitude of the charge on each plate
correct check mark nC
(c) the stored energy
correct check mark µJ
(d) the electric field between the plates
wrong check mark V/m
(e) the energy density between the plates
J/m3

So I figured because the capacitor is a conductor, i know that the e-field just outside a conductor is E = Area Charge Density/Eo or
E = (Q/A)/Eo;
E = (7.57nC)/(.0038cm^2)/8.85e-12 = 225096639 N/C.
Which was wrong, any ideas why? Thanks.

 

[vanesch]

Hello everyone. I can't get the 2nd to last question on this problem...
It says:
A parallel-plate air-filled capacitor having area 38 cm2 and plate spacing 2.0 mm is charged to a potential difference of 450 V. Find the following values.
(a) the capacitance
correct check mark pF
(b) the magnitude of the charge on each plate
correct check mark nC
(c) the stored energy
correct check mark µJ
(d) the electric field between the plates
wrong check mark V/m
(e) the energy density between the plates
J/m3

So I figured because the capacitor is a conductor, i know that the e-field just outside a conductor is E = Area Charge Density/Eo or
E = (Q/A)/Eo;
E = (7.57nC)/(.0038cm^2)/8.85e-12 = 225096639 N/C.
Which was wrong, any ideas why? Thanks.

I didn't check the numbers but the approach should be (about) right.
However, you're making life difficult: potential is E-field integrated over distance. Now in a parallel plate cap, the E-field is supposed to be constant, so this reduces to V = E x distance. You know the potential difference over the plates and their distance...

 

[mr_coffee]

THank you so much! that worked great and was much easier, i must have messed up in the calculations!